Problem: Divide the following complex numbers. $ \dfrac{-2+16i}{-1-5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1+5i}$ $ \dfrac{-2+16i}{-1-5i} = \dfrac{-2+16i}{-1-5i} \cdot \dfrac{{-1+5i}}{{-1+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2+16i) \cdot (-1+5i)} {(-1-5i) \cdot (-1+5i)} = \dfrac{(-2+16i) \cdot (-1+5i)} {(-1)^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2+16i) \cdot (-1+5i)} {(-1)^2 - (-5i)^2} = $ $ \dfrac{(-2+16i) \cdot (-1+5i)} {1 + 25} = $ $ \dfrac{(-2+16i) \cdot (-1+5i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2+16i}) \cdot ({-1+5i})} {26} = $ $ \dfrac{{-2} \cdot {(-1)} + {16} \cdot {(-1) i} + {-2} \cdot {5 i} + {16} \cdot {5 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{2 - 16i - 10i + 80 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{2 - 16i - 10i - 80} {26} = \dfrac{-78 - 26i} {26} = -3-i $